course Mth 151
********************************************* Your solution: A= 237,864,6,3972 at least one even digit A’= 397,135,1,9937 no even digits B= {3,8,35,89,104,357,43213 A and B in common = 8,89,104,4321. These are all in A & B with one even digit common to both collections. B and A’ have the numbers that do not have at least one even number = 3,35,257
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Given Solution: `aOf the numbers in B, 8, 89, 104, 4321 each have at least one even digit and so are common to both sets. 3 is odd, both of the digits in the number 35 are odd, as are all three digits in the number 357. Both of these numbers are therefore in A ' . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The 2nd question gave me a little trouble. I could use some help. Is it asking for all of the numbers that are in common but does not have one even digit. Self-critique Rating: ********************************************* Question: `q002. I have in a room 8 people with dark hair brown, 2 people with bright red hair, and 9 people with light brown or blonde hair. Nobody has more than one hair color. Is it possible that there are exactly 17 people in the room?
********************************************* Your solution: 8= dark hair brown 2= red hair 9=light My answer is no, because you have brown, light brown red and blonde hair and you add together there are 8+2+9=19 can’t have less.
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Given Solution: `aIf we assume that dark brown, light brown or blonde, and bright red hair are mutually exclusive (i.e., someone can't be both one category and another, much less all three), then we have at least 8 + 2 + 9 = 19 people in the room, and it is not possible that we have exactly 17. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Althought I got this one right, in the world today, the kids do have two different hair colors.
********************************************* Your solution: 6= dark hair 10= blue eyes 14= all together This is something that one person can have both of, so there could be sixteen in this collection. A person may be counted twice in this case.
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Given Solution: `aThe key here is that there is nothing mutully exclusive about these categories-a person can have blue eyes as well as dark hair. So if there are 2 people in the room who have dark hair and blue eyes, which is certainly possible, then when we add 10 + 6 = 16 those two people would be counted twice, once among the 6 blue-eyed people and once among the 10 dark-haired people. So the 16 we get would be 2 too high. To get the correct number we would have to subtract the 2 people who were counted twice to get 16 - 2 = 14 people. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 2 ********************************************* Question: `q004. In a set of 100 child's blocks 60 blocks are cubical and 40 blocks are cylindrical. 30 of the blocks are red and 20 of the red blocks are cubical. How many of the cylindrical blocks are red?
********************************************* Your solution: 100= child’s blocks 60= cubical 40= cylindrical 30= red 20= red cubicals There are 30 red blocks 20 are cubical this leaves only 10 red blocks left so 10 are red cylindrical.
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Given Solution: `aOf the 30 red blocks 20 are cubical, so the rest must be cylindrical. This leaves 10 red cylindrical blocks. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I just subtracted 20 from 30 left only 10 Self-critique Rating: 3